A ball is projected with kinetic energy E, at an angle of 60° to the horizontal. The kinetic energy of this ball at the highest point of its flight will become:

Option 1 -  zero

Option 2 - <math> <mrow> <mfrac> <mrow> <mi>E</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math>

Option 3 - <math> <mrow> <mfrac> <mrow> <mi>E</mi> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> </mrow> </math>

Option 4 - E

4 Views|Posted 6 months ago

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Answered by

Vishal Baghel | Contributor-Level 10

6 months ago

Correct Option - 3

Detailed Solution:

$E = \frac{1}{2} m u^{2}$

$E_{H i g h e s t p o i n t} = \frac{1}{2} m {(\frac{u}{2})}^{2} = \frac{1}{8} m u^{2} = \frac{E}{4}$

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